What is the integral of 1/(x-2x^{3/4)-8sqrt(x)} ?

The integral of 1/(x-2x^{3/4)-8sqrt(x)} is 4(1/3 ln|\sqrt[4]{x}+2|+2/3 ln|\sqrt[4]{x}-4|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x-2x^{3/4)-8sqrt(x)}

\int \frac{1}{x - 2 x ^{\frac{3}{4}} - 8 x} d x

Solution steps

\int \frac{1}{x - 2 x ^{\frac{3}{4}} - 8 x} d x

Apply u-substitution

: \int \frac{4 u}{u ^{2} - 2 u - 8} d u

= \int \frac{4 u}{u ^{2} - 2 u - 8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{u}{u ^{2} - 2 u - 8} d u

Take the partial fraction of

\frac{u}{u ^{2} - 2 u - 8} : \frac{1}{3 ( u + 2 )} + \frac{2}{3 ( u - 4 )}

= 4 \cdot \int \frac{1}{3 ( u + 2 )} + \frac{2}{3 ( u - 4 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int \frac{1}{3 ( u + 2 )} d u + \int \frac{2}{3 ( u - 4 )} d u)

\int \frac{1}{3 ( u + 2 )} d u = \frac{1}{3} ln ∣ u + 2 ∣

\int \frac{2}{3 ( u - 4 )} d u = \frac{2}{3} ln ∣ u - 4 ∣

= 4 (\frac{1}{3} ln ∣ u + 2 ∣ + \frac{2}{3} ln ∣ u - 4 ∣)

Substitute back

u = x^{\frac{1}{4}}

= 4 (\frac{1}{3} ln x^{\frac{1}{4}} + 2 + \frac{2}{3} ln x^{\frac{1}{4}} - 4)

Apply radical rule:

Add a constant to the solution

n = 0 \sum \infty \frac{5 + n}{1 0 ^{n}}

\int (\frac{csc ( x ) + cot ( x )}{sin ( x )}) d x

\int x^{4035} e^{x^{2018}} d x

x \to 1 lim (x arcsin (\frac{x}{2}))

\frac{d}{d x} (3 \cdot 1 0^{2 x})

What is the integral of 1/(x-2x^{3/4)-8sqrt(x)} ?
The integral of 1/(x-2x^{3/4)-8sqrt(x)} is 4(1/3 ln|\sqrt[4]{x}+2|+2/3 ln|\sqrt[4]{x}-4|)+C